**Rotation of Axes (Change of Direction)**

**1) **When the axes rotated through an angle 60, then new co-ordinates of three points are the following (3, 4).

**Solution: **Given that,

New co-ordinates are (3, 4)

X = 3, Y = 4

x = X cos θ – Y sin θ

= 3 cos60 – 4 sin60

= 3 (½) – 4 \(\frac{\sqrt{3}}{2}\),

= \(\frac{3-4\sqrt{3}}{2}\).

y = X sin θ + Y cos θ

= 3 sin60 + 4 cos60

= 3 \(\frac{\sqrt{3}}{2}\) + 4 (½)

= \(\frac{4+\sqrt{3}}{2}\).

Co-ordinate of P are \(\left( \frac{3-4\sqrt{3}}{2},\frac{4+\sqrt{3}}{2} \right)\).

**2)** Find the angle through which the axes are to be rotated so as to remove the xy term in the equation x² + 4xy + y² – 2x + 2y – 6 = 0.

**Solution: **Comparing the equation

x² + 4xy + y² – 2x + 2y – 6 = 0 with ax² + 2hxy + by² + 2gx + 2fy + c = 0

a = 1, h = 2, b = 1, g = -1, f = 1, c = -6

let θ be the angle of rotation of axes, then θ = ½ tan⁻¹\(\left( \frac{2h}{a-b} \right)\),

= ½ tan⁻¹ (4/ 1 – 1) = ½ tan⁻¹ (4/0)

= ½ tan⁻¹(∞) = ½ x π/2

θ = π/4

**3)** When the axes are rotated through an angle 45, the transformed equation of a curve is 17x² – 16xy + 17y² = 255. Find the original equation of the curve?

**Solution: **Angle of rotation = θ = 45

X = x cosθ + y sinθ

= x cos45 + y sin45 = (x + y)/2

Y = -x sinθ + y cosθ

= -xsin45 + ycos45 = (-x + y)/2

The original equation is 17x² – 16xy + 17y² = 255

⇒ \({{17\left( \frac{x+y}{2} \right)}^{2}}-16\left( \frac{x+y}{2} \right)\left( \frac{-x+y}{2} \right)+17{{\left( \frac{-x+y}{2} \right)}^{2}}=225\),

⇒ \(17\left( \frac{{{x}^{2}}+{{y}^{2}}+2xy}{2} \right)-16\left( \frac{{{x}^{2}}-{{y}^{2}}}{2} \right)+17\left( \frac{{{x}^{2}}+{{y}^{2}}+2xy}{2} \right)=225\),

⇒ 17 (x² + y²) – 8 (x² – y²) = 225

⇒ 19 x² + 25 y² = 225.